chaseashley24
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1.State whether the relationship between the variables in the table is a direct variation, an inverse variation, or neither. If it is a direct or inverse variation, write a function to model it.

2.How do the graphs of y=1/x and y=/5x+6

compare?

1.State whether the relationship between the variables in the table is a direct variation, an inverse variation, or neither. If it is a direct or inverse variat class=
1.State whether the relationship between the variables in the table is a direct variation, an inverse variation, or neither. If it is a direct or inverse variat class=

Answer :

jcherry99

Answer:


Step-by-step explanation:

Left Pane

Argument

[Left Graph]: The table should be graphed to begin with. Some people don't like that idea very much, but it is a quick way to get an answer. I  took the liberty of adding 0,0

When a line goes through 0,0 the variation in this case is a direct variation. Then general equation for a direct variation is y = k*x

K is the slope. You should get 3.

m = (y2 - y1) / (x2 - x1)

Givens

  • y2 = 60
  • y1 = 9
  • x2 = 20
  • x1 = 3

Solution

  • m = (60 - 9) / (20 - 3)
  • m =  51 / 17
  • m = 3

Answer

Direct variation

Right Panel

Argument

The best way to show this is to (again) graph it. [Right Graph]

  • Red: y = 1/x
  • Blue: y = 5/(x + 6)
  • Green: x = 0
  • Orange x = - 6

The last two are asymptote where if the denominator approaches these values, the hyperbola goes upwards depending on what part of the graph being observed.

The 6 has the property of moving the graph left (x + 6). And the 5 has the property of stretching the graph up and down. Note the word used is stretched not moved.

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